What is the integral of x/((1+x)^2) ?

The integral of x/((1+x)^2) is ln|1+x|+1/(1+x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/((1+x)^2)

integral

\frac{x}{( 1 + x ) ^{2}}

ln ∣ 1 + x ∣ + \frac{1}{1 + x} + C

Solution steps

\int \frac{x}{( 1 + x ) ^{2}} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{2}} d u

= \int \frac{u - 1}{u ^{2}} d u

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= \int \frac{1}{u} - \frac{1}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= ln ∣ u ∣ - (- \frac{1}{u})

Substitute back

u = 1 + x

= ln ∣ 1 + x ∣ - (- \frac{1}{1 + x})

Simplify

= ln ∣ 1 + x ∣ + \frac{1}{1 + x}

Add a constant to the solution

= ln ∣ 1 + x ∣ + \frac{1}{1 + x} + C

\frac{1}{e ^{- x}}

- tan^{2} (x)

\frac{1}{sec ( x )}

\frac{1}{( x ^{2} - x )}

\frac{x ^{3}}{x}

What is the integral of x/((1+x)^2) ?
The integral of x/((1+x)^2) is ln|1+x|+1/(1+x)+C